Problem: Let $h(x)=\dfrac{e^{2x}}{x-3}$ Where does $h$ have critical points? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=0$ (Choice B) B $x=3$ (Choice C) C $x=3.5$ (Choice D) D $h$ has no critical points.
Explanation: A critical point of $h$ is a point in the domain of $h$ where the derivative is either equal to zero or undefined. So in order to find the critical points of $h$, let's find its derivative. $\begin{aligned} h'(x)&=\dfrac{d}{dx}\left[ \dfrac{e^{2x}}{x-3} \right] \\\\ &=\dfrac{(x-3)\cdot \dfrac{d}{dx}[e^{2x}]-e^{2x}\cdot \dfrac{d}{dx}[x-3]}{(x-3)^2} \\\\ &=\dfrac{(x-3)\cdot 2e^{2x}-e^{2x}\cdot 1}{(x-3)^2} \\\\ &=\dfrac{e^{2x}(2x-7)}{(x-3)^2} \end{aligned}$ Now let's look for $x$ -values where $h'$ is zero or undefined. $\dfrac{e^{2x}(2x-7)}{(x-3)^2}=0$ at $x=3.5$. $\dfrac{e^{2x}(2x-7)}{(x-3)^2}$ is undefined at $x=3$. However, $h(x)=\dfrac{e^{2x}}{x-3}$ is also undefined at $x=3$ so it isn't a critical point. In conclusion, this is the only $x$ -value where $h$ has a critical point: $x=3.5$